Is the function given below continuous/differentiable at $x=3$ ? $g(x)=\begin{cases} -3x+2&,&x<3 \\\\ x^2-5x-1&,&x\geq3 \end{cases}$ Choose 1 answer: Choose 1 answer: (Choice A) A Continuous but not differentiable (Choice B) B Differentiable but not continuous (Choice C) C Both continuous and differentiable (Choice D) D Neither continuous nor differentiable
Checking for continuity at $x=3$ For the function to be continuous at $x=3$, we need the two-sided limit $\lim_{x\to 3}g(x)$ to exist and be equal to $g(3)$. This is the same as requiring that the two one-sided limits $\lim_{x\to 3^-}g(x)$ and $\lim_{x\to 3^+}g(x)$ exist and are equal to $g(3)$. According to $g$ 's definition, $g(3)=(3)^2-5(3)-1=-7$. $\lim_{x\to 3^-}g(x)$ $-3x+2$ evaluated at $x=3$ is equal to $-7$. Since $-3x+2$ is continuous, we can be certain that $\lim_{x\to 3^-}g(x)=-7$. $\lim_{x\to 3^+}g(x)$ $x^2-5x-1$ evaluated at $x=3$ is equal to $-7$. Since $x^2-5x-1$ is continuous, we can be certain that $\lim_{x\to 3^+}g(x)=-7$. We saw that the two one-sided limits exist and are equal to $g(3)$, so the function is continuous at $x=3$. Checking for differentiability at $x=3$ For the function to be differentiable at $x=3$, we need the two-sided limit $\lim_{x\to 3}\dfrac{g(x)-g(3)}{x-3}=\lim_{x\to 3}\dfrac{g(x)-(-7)}{x-3}$ to exist. This is the same as requiring that the two one-sided limits $\lim_{x\to 3^-}\dfrac{g(x)-(-7)}{x-3}$ and $\lim_{x\to 3^+}\dfrac{g(x)-(-7)}{x-3}$ exist and have the same value. $\lim_{x\to 3^-}\dfrac{g(x)-(-7)}{x-3}=-3$ $\lim_{x\to 3^+}\dfrac{g(x)-(-7)}{x-3}=1$ The two limits exist, but they are not equal. Therefore, the function is not differentiable at $x=3$. Graphically, the function has a sharp turn at this point. [I would like to see that, please!] In conclusion, the function is continuous but not differentiable at $x=3$.